Phenol Adsorption Calculation
This solution calculates the adsorption capacity of a dry adsorbent for phenol based on changes in solution concentration.
Key Data Provided
- Mass of dry adsorbent: $500$ mg = $0.5$ g
- Volume of phenol solution: $100$ mL = $0.1$ L
- Initial phenol concentration: $100$ mg/L
- Residual (final) phenol concentration: $30$ mg/L
Calculating Phenol Removed
1. Calculate Initial Phenol Mass:
The total mass of phenol initially in the solution is:
$ m_{\text{initial}} = \text{Volume} \times \text{Initial Concentration} $
$ m_{\text{initial}} = 0.1 \text{ L} \times 100 \frac{\text{mg}}{\text{L}} = 10 \text{ mg} $
2. Calculate Final Phenol Mass:
The total mass of phenol remaining in the solution after adsorption is:
$ m_{\text{final}} = \text{Volume} \times \text{Final Concentration} $
$ m_{\text{final}} = 0.1 \text{ L} \times 30 \frac{\text{mg}}{\text{L}} = 3 \text{ mg} $
3. Calculate Mass Adsorbed:
The mass of phenol adsorbed by the adsorbent is the difference between the initial and final masses:
$ m_{\text{adsorbed}} = m_{\text{initial}} - m_{\text{final}} $
$ m_{\text{adsorbed}} = 10 \text{ mg} - 3 \text{ mg} = 7 \text{ mg} $
Determining Adsorption Capacity
The adsorption capacity is defined as the mass of solute adsorbed per unit mass of adsorbent.
Adsorption Capacity = $\frac{\text{Mass of Phenol Adsorbed}}{\text{Mass of Adsorbent}}$
Capacity = $\frac{7 \text{ mg}}{0.5 \text{ g}} = 14 \frac{\text{mg}}{\text{g}}$
The calculated adsorption capacity is $14$ mg/g.
Important Questions from Adsorption
- A volatile organic compound (VOC) is to be adsorbed from air onto a bed of activated carbon. The equilibrium capacity of activated carbon at the feed conditions is $0.4 \text{ grams}$ VOC per gram of activated carbon. The column contains $4 \text{ grams}$ of activated carbon per $\text{cm}^2$ of cross-section. The feed rate into the adsorber column is $0.2 \text{ grams}$ VOC $\text{cm}^{-2}\text{ h}^{-1}$. Breakthrough time is defined as the time at which the concentration at the exit of the bed ($c$) reaches a value of $0.05c_0$, where $c_0$ is the concentration of VOC in the feed. The breakthrough time for the bed is $2.1 \text{ h}$. The area under $c / c_0$ curve between the initial and breakthrough times is $0.1 \text{ h}$. Which one of the following is the fraction of unused bed at breakthrough?
- The correct statement(s) for a surface catalyzed unimolecular gaseous reaction is(are)
- Adsorption on activated carbon is to be used for reducing phenol concentration in wastewater from 0.04 mol/l to 0.008 mol/l. The adsorption isotherm at the operating temperature can be expressed as $q = 0.025C^{1/3}$; where $q$ is the phenol concentration in solid (mol/g solid) and $C$ is the phenol concentration in water (mol/l). The minimum amount of solid (in grams) required per liter of wastewater (up to one decimal place) is ________
- At 30°C, the amounts of acetone adsorbed at partial pressures of 10 and 100 mmHg are 0.1 and 0.4 kg acetone/kg activated carbon, respectively. Assume Langmuir isotherm describes the adsorption of acetone on activated carbon. What is the amount of acetone adsorbed (in kg per kg of activated carbon) at a partial pressure of 50 mmHg and 30°C?
At 273 K and 10 bar, the Langmuir adsorption of a gas on a solid surface gave the fraction of surface coverage as 0.01. The Langmuir adsorption isotherm constant is ________ bar$^{-1}$. (Give the answer to the third decimal place)
